We will proceed by induction on the number of edges of \(F\) . If \(F\)  has no edges and its vertex weights under \(\upsilon \)  are given in weakly decreasing order by the partition \(\alpha \) , then the chromatic symmetric function of \((F,\upsilon )\)  is exactly the power sum symmetric function \(p_\alpha \) , e.g. by [1]}. Moreover, since \(F\)  has no edges, the combinatorial Laplacian \(L_F\)  is just the zero matrix. Therefore,
\(P_{(F,\upsilon )}(t) = \det (t W_\upsilon )=\left(\prod _{v\in V}\upsilon (v)\right)t^{|F|} = \left(\prod _{i=1}^{\ell (\alpha )}\alpha _i\right)t^{\ell (\alpha )}=\varphi (X_{(F,\upsilon )}).\) 
For the inductive step, assume \(F\)  has at least one edge \(e\) . It is not a loop, as \(F\)  is a forest. Since \((F-e,\upsilon )\)  and \((F/e,\upsilon /e)\)  are both integer vertex-weighted forests with strictly fewer edges, we have by the inductive hypothesis and the deletion-contraction relations in Theorem REF  (recalling that for vertex-weighted graphs we treat all edge weights as being 1) and Theorem REF  that
\(P_{(F,\upsilon )}(t) = P_{(F-e,\upsilon )}(t)-P_{(F/e,\upsilon /e)}(t) = \varphi (X_{(F-e,\upsilon )})-\varphi (X_{(F/e,\upsilon /e)}) = \varphi (X_{(F,\upsilon )}),\) 
as claimed.
